Chapter 28 Magnetic Fields

28.2: What Produces Magnetic Field?: One way that magnetic fields are produced is to use moving electrically charged particles, such as a current in a wire, to make an electromagnet. The current produces a magnetic field that is utilizable. The other way to produce a magnetic field is by means of elementary particles such as electrons, because these particles have an intrinsic magnetic field around them. The magnetic fields of the electrons in certain materials add together to give a net magnetic field around the material. Such addition is the reason why a permanent magnet, has a permanent magnetic field. In other materials, the magnetic fields of the electrons cancel out, giving no net magnetic field surrounding the material.

28.3: The Definition of B: We can define a magnetic field, B, by firing a charged particle through the point at which is to be defined, using various directions and speeds for the particle and determining the force that acts on the particle at that point. B is then defined to be a vector quantity that is directed along the zero-force axis. The magnetic force on the charged particle, F B, is defined to be: Here q is the charge of the particle, v is its velocity, and B the magnetic field in the region. The magnitude of this force is then: Here  is the angle between vectors v and B.

28.3: Finding the Magnetic Force on a Particle:

28.3: The Definition of B: The SI unit for B that follows is newton per coulomb-meter per second. For convenience, this is called the tesla (T): An earlier (non-SI) unit for B is the gauss (G), and

28.3: Magnetic Field Lines:  The direction of the tangent to a magnetic field line at any point gives the direction of B at that point.  The spacing of the lines represents the magnitude of B —the magnetic field is stronger where the lines are closer together, and conversely.

Example, Magnetic Force on a Moving Charged Particle :

28.4: Crossed Fields, Discovery of an Electron: When the two fields in Fig are adjusted so that the two deflecting forces acting on the charged particle cancel, we have Thus, the crossed fields allow us to measure the speed of the charged particles passing through them. The deflection of a charged particle, moving through an electric field, E, between two plates, at the far end of the plates (in the previous problem) is Here, v is the particle’s speed, m its mass, q its charge, and L is the length of the plates.

28.5: Crossed Fields, The Hall Effect: Fig A strip of copper carrying a current i is immersed in a magnetic field. (a)The situation immediately after the magnetic field is turned on. The curved path that will then be taken by an electron is shown. (b) The situation at equilibrium, which quickly follows. Note that negative charges pile up on the right side of the strip, leaving uncompensated positive charges on the left. Thus, the left side is at a higher potential than the right side. (c) For the same current direction, if the charge carriers were positively charged, they would pile up on the right side, and the right side would be at the higher potential. A Hall potential difference V is associated with the electric field across strip width d, and the magnitude of that potential difference is V =Ed. When the electric and magnetic forces are in balance (Fig. 28-8b), where v d is the drift speed. But, Where J is the current density, A the cross-sectional area, e the electronic charge, and n the number of charges per unit volume. Therefore, Here, l=( A/d), the thickness of the strip.

Example, Potential Difference Setup Across a Moving Conductor:

Example, Potential Difference Setup Across a Moving Conductor, cont.:

28.6: A Circulating Charged Particle: Consider a particle of charge magnitude |q| and mass m moving perpendicular to a uniform magnetic field B, at speed v. The magnetic force continuously deflects the particle, and since B and v are always perpendicular to each other, this deflection causes the particle to follow a circular path. The magnetic force acting on the particle has a magnitude of |q|vB. For uniform circular motion Fig Electrons circulating in a chamber containing gas at low pressure (their path is the glowing circle). A uniform magnetic field, B, pointing directly out of the plane of the page, fills the chamber. Note the radially directed magnetic force F B ; for circular motion to occur, F B must point toward the center of the circle, (Courtesy John Le P.Webb, Sussex University, England)

28.6: Helical Paths: Fig (a) A charged particle moves in a uniform magnetic field, the particle’s velocity v making an angle f with the field direction. (b) The particle follows a helical path of radius r and pitch p. (c) A charged particle spiraling in a nonuniform magnetic field. (The particle can become trapped, spiraling back and forth between the strong field regions at either end.) Note that the magnetic force vectors at the left and right sides have a component pointing toward the center of the figure. The velocity vector, v, of such a particle resolved into two components, one parallel to and one perpendicular to it: The parallel component determines the pitch p of the helix (the distance between adjacent turns (Fig b)). The perpendicular component determines the radius of the helix. The more closely spaced field lines at the left and right sides indicate that the magnetic field is stronger there. When the field at an end is strong enough, the particle “reflects” from that end. If the particle reflects from both ends, it is said to be trapped in a magnetic bottle.

Example, Helical Motion of a Charged Particle in a Magnetic Field:

Example, Uniform Circular Motion of a Charged Particle in a Magnetic Field:

28.7: Cyclotrons : Suppose that a proton, injected by source S at the center of the cyclotron in Fig , initially moves toward a negatively charged dee. It will accelerate toward this dee and enter it. Once inside, it is shielded from electric fields by the copper walls of the dee; that is, the electric field does not enter the dee. The magnetic field, however, is not screened by the (nonmagnetic) copper dee, so the proton moves in a circular path whose radius, which depends on its speed, is given by (r =mv/|q|B). Let us assume that at the instant the proton emerges into the center gap from the first dee, the potential difference between the dees is reversed. Thus, the proton again faces a negatively charged dee and is again accelerated. This process continues, the circulating proton always being in step with the oscillations of the dee potential, until the proton has spiraled out to the edge of the dee system. There a deflector plate sends it out through a portal. The frequency f at which the proton circulates in the magnetic field (and that does not depend on its speed) must be equal to the fixed frequency f osc of the electrical oscillator:

28.7: The Proton Synchrotron : At proton energies above 50 MeV, the conventional cyclotron begins to fail. Also, for a 500 GeV proton in a magnetic field of 1.5 T, the path radius is 1.1 km. The corresponding magnet for a conventional cyclotron of the proper size would be impossibly expensive. In the proton synchrotron the magnetic field B, and the oscillator frequency f osc, instead of having fixed values as in the conventional cyclotron, are made to vary with time during the accelerating cycle. When this is done properly, (1)the frequency of the circulating protons remains in step with the oscillator at all times, and (2)the protons follow a circular—not a spiral—path. Thus, the magnet need extend only along that circular path, not over some 4x10 6 m 2. The circular path, however, still must be large if high energies are to be achieved. The proton synchrotron at the Fermi National Accelerator Laboratory (Fermilab) in Illinois has a circumference of 6.3 km and can produce protons with energies of about 1 TeV ( eV).

Example, Accelerating a Charged Particle in a Synchrotron:

28.8: Magnetic Force on a Current-Carrying Wire:

28.8: Magnetic Force on a Current-Carrying Wire: Consider a length L of the wire in the figure. All the conduction electrons in this section of wire will drift past plane xx in a time t =L/vd. Thus, in that time a charge will pass through that plane that is given by Here L is a length vector that has magnitude L and is directed along the wire segment in the direction of the (conventional) current. If a wire is not straight or the field is not uniform, we can imagine the wire broken up into small straight segments. The force on the wire as a whole is then the vector sum of all the forces on the segments that make it up. In the differential limit, we can write and we can find the resultant force on any given arrangement of currents by integrating Eq over that arrangement.

Example, Magnetic Force on a Wire Carrying Current:

28.9: Torque on a Current Loop: The two magnetic forces F and –F produce a torque on the loop, tending to rotate it about its central axis.

28.9: Torque on a Current Loop: To define the orientation of the loop in the magnetic field, we use a normal vector n that is perpendicular to the plane of the loop. Figure 28-19b shows a right-hand rule for finding the direction of n. In Fig c, the normal vector of the loop is shown at an arbitrary angle  to the direction of the magnetic field. For side 2 the magnitude of the force acting on this side is F 2 =ibB sin(90°-  )=ibB cos  =F 4. F 2 and F 4 cancel out exactly. Forces F 1 and F 3 have the common magnitude iaB. As Fig c shows, these two forces do not share the same line of action; so they produce a net torque. For N loops, when A=ab, the area of the loop, the total torque is:

28.10: The Magnetic Dipole Moment,  : Definition: Here, N is the number of turns in the coil, i is the current through the coil, and A is the area enclosed by each turn of the coil. Direction: The direction of  is that of the normal vector to the plane of the coil. The definition of torque can be rewritten as: Just as in the electric case, the magnetic dipole in an external magnetic field has an energy that depends on the dipole’s orientation in the field: A magnetic dipole has its lowest energy (-  B cos 0=-  B) when its dipole moment  is lined up with the magnetic field. It has its highest energy (-  B cos 180°=+  B) when  is directed opposite the field.

28.10: The Magnetic Dipole Moment,  : From the above equations, one can see thatthe unit of  can be the joule per tesla (J/T), or the ampere–square meter.